A 12kg wad of clay is hurled horizontally at a 100g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50m before coming to rest. If the coefficient of friction between the bloack and the surface is .650, what is the SPEED of the clay immediately before impact?

2. Relevant equations

This is a perfectly inelastic collision correct?

My book tells me in the answer section to use the following formula, but Im not sure where it came from or why to use this. Am I to use this for equations related to inelastic collisions and friction involved?

1/2(m1+m2)V2^2 = friction(m1+m2)gL

How did it come up with that???? I guess I am just having trouble figuring out how I was suppose to come up with that formula.

Staff: Mentor

My book tells me in the answer section to use the following formula, but Im not sure where it came from or why to use this. Am I to use this for equations related to inelastic collisions and friction involved?

1/2(m1+m2)V2^2 = friction(m1+m2)gL

How did it come up with that???? I guess I am just having trouble figuring out how I was suppose to come up with that formula.

It's telling you that the kinetic energy the system had after the collision must equal the work done by friction. Note that V2 is the speed of the 'clay+block' after the collision. You'll use that to figure out the speed of the clay before the collision.

Staff: Mentor

Ok, so first thing I do when looking at this is figure out the Vf of the system after the collision.

Right. You'll do that by solving the second part first, since you have all the needed data. (It's often the case that its best to work a problem 'backwards'.)

With the formula I supplied for the first step Vf = m1V1i + m2V2i / (m1 + m2) , how do I go from there, to here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL

You'll solve the second step first.

How will you relate the two steps? Hint: How does Vf in the first formula relate to V2 in the second? (Generally it's a good idea to use the same symbols for the same quantities. Otherwise things get confusing.)

Staff: Mentor

You ask what is V2 ? Well, thats the Velocity of the wood, which is zero, correct? What significance is that may I ask?

Yes, in your conservation of momentum equation (which describes the collision), V2 stands for the initial speed of the block, which is zero. You'll need that equation to solve for the initial speed of the bullet. (You'll use this equation last.)

EDIT: I guess what im missing is how I would come to the formula to get the speed of the system, as detailed here: 1/2(m1+m2)V2^2 = friction(m1+m2)gL

That's just an energy equation stating that the change in KE of the system after the collision is equal to the work done by friction. What does V2 stand for here?